Why would NaOH solution need to be added when preparing the K2CrO4 stock solution

Because the second ionization of H2CrO4 is not 100%; i.e., it is a weak acid.

H2CrO4 ==> H^+ + HCrO4^-
HCrO4^- ==> H^+ + CrO^2-
k1 = 100%
k2 = about 5E-7 in my set of tables.
When CrO4^2- is added to water it hydrolyzes to form an equilibrium between CrO4^2- and HCrO4^-. Adding base keeps essentially 100% or the Cr in the form of CrO4^2-.